How do you find #\lim _ { x \rightarrow - 14} \frac { 13- \sqrt { x ^ { 2} - 27} } { x + 14}#?

Question

Evelyn Agard · Accepted Answer

# 14/13.#

#"The Reqd. Limit L="lim_(x to -14) (13-sqrt(x^2-27))/(x+14),#

#=lim_(x to -14)(13-sqrt(x^2-27))/(x+14)xx(13+sqrt(x^2-27))/(13+sqrt(x^2-27)),#

#=lim_(x to -14){13^2-(x^2-27)}/{(x+14)(13+sqrt(x^2-27)),#

#=lim_(x to -14)(169+27-x^2)/{(x+14)(13+sqrt(x^2-27)),#

#=lim_(x to -14)(196-x^2)/{(x+14)(13+sqrt(x^2-27)),#

#=lim_(x to -14){cancel((14+x))(14-x)}/{cancel((x+14))(13+sqrt(x^2-27)),#

#={14-(-14)}/{13+sqrt((-14)^2-27)},#

#=28/(13+sqrt(196-27)),#

#=28/(13+sqrt169),#

#=28/(13+13)=28/26,#

#rArr L=14/13.#

Enjoy Maths.!

Ezra Adams · Answer

# 14/13.#

Let us solve the Problem, using the following Standard Form :

#lim_(x to a) (x^n-a^n)/(x-a)=n*a^(n-1)," equivalently, "#

#lim_(x to a)(a^n-x^n)/(a-x)=n*a^(n-1)..............(star).#

We suppose,

#y=x^2-27," so that, as "x to -14, y to (-14)^2-27, or, y to 169.#

#:."The Reqd. Lim.="lim_(x to -14) (13-sqrt(x^2-27))/(x+14),#

#=lim_(y to 169)(169^(1/2)-y^(1/2))/(169-y)xx(169-y)/(x+14),#

#={1/2*169^(1/2-1)}{lim_(x to -14)(169-(x^2-27))/(x+14)}...[because,(star)],#

#={1/2*(13^2)^(-1/2)}{lim_(x to -14)(196-x^2)/(14+x)},#

#=(1/2*13^(-1)){-lim_(x to -14)(x^2-(-14)^2)/(x-(-14))},#

#=(1/2*1/13)(-2*(-14)^(2-1))............[because,(star)],#

#=1/2*1/13*(-2*(-14)).#

#rArr "The Reqd. Lim.=" 14/13,# as before!

Enjoy Maths.!

Christopher Allers · Answer

undefined To find $\lim_{x \to -14} \frac{13 - \sqrt{x^2 - 27}}{x + 14}$, we can rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator. The conjugate of $13 - \sqrt{x^2 - 27}$ is $13 + \sqrt{x^2 - 27}$.

After rationalizing, we simplify the expression and then substitute $x = -14$ into the resulting expression to find the limit.

The result of this process is that the limit is $\frac{1}{14}$.